But you never _proved_ the inequality a < p, so you don't get to use it.

Moreover, the equation

a^p + b^p = c^p

with the restrictions

a,b,c positive integers

p prime

does not imply min(a,b) < p.

To see this, just use p = 2 with a,b,c = 3,4,5.

You tried to argue that you can't have p=2 since the inequality min(a,b) < p would then force min(a,b) = 1,leading to an easy contradiction. But you can't use the inequality min(a,b) < p without proving it, and the example p = 2 with a,b,c = 3,4,5 makes it clear that you can't prove it.